What is the integral of cos(at)e^{-st} ?

The integral of cos(at)e^{-st} is (ae^{-st}sin(at))/(a^2+s^2)-(se^{-st}cos(at))/(a^2+s^2)+C

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integral of cos(at)e^{-st}

\int cos (a t) e^{- s t} d t

\frac{a e ^{- s t} sin ( a t )}{a ^{2} + s ^{2}} - \frac{s e ^{- s t} cos ( a t )}{a ^{2} + s ^{2}} + C

Solution steps

\int cos (a t) e^{- s t} d t

Apply Integration By Parts

= e^{- s t} \frac{1}{a} sin (a t) - \int - s e^{- s t} \frac{1}{a} sin (a t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{- s t} \frac{1}{a} sin (a t) - (- s \frac{1}{a} \cdot \int e^{- s t} sin (a t) d t)

Simplify

- s \frac{1}{a} : - \frac{s}{a}

= e^{- s t} \frac{1}{a} sin (a t) - (- \frac{s}{a} \cdot \int e^{- s t} sin (a t) d t)

Apply Integration By Parts

= e^{- s t} \frac{1}{a} sin (a t) - (- \frac{s}{a} (- e^{- s t} \frac{1}{a} cos (a t) - \int s e^{- s t} \frac{1}{a} cos (a t) d t))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{- s t} \frac{1}{a} sin (a t) - (- \frac{s}{a} (- e^{- s t} \frac{1}{a} cos (a t) - s \frac{1}{a} \cdot \int e^{- s t} cos (a t) d t))

Simplify

s \frac{1}{a} : \frac{s}{a}

= e^{- s t} \frac{1}{a} sin (a t) - (- \frac{s}{a} (- e^{- s t} \frac{1}{a} cos (a t) - \frac{s}{a} \cdot \int e^{- s t} cos (a t) d t))

Therefore

\int e^{- s t} cos (a t) d t = e^{- s t} \frac{1}{a} sin (a t) - (- \frac{s}{a} (- e^{- s t} \frac{1}{a} cos (a t) - \frac{s}{a} \cdot \int e^{- s t} cos (a t) d t))

Isolate

\int e^{- s t} cos (a t) d t

= \frac{a e ^{- s t} sin ( a t )}{a ^{2} + s ^{2}} - \frac{s e ^{- s t} cos ( a t )}{a ^{2} + s ^{2}}

Add a constant to the solution

= \frac{a e ^{- s t} sin ( a t )}{a ^{2} + s ^{2}} - \frac{s e ^{- s t} cos ( a t )}{a ^{2} + s ^{2}} + C

What is the integral of cos(at)e^{-st} ?
The integral of cos(at)e^{-st} is (ae^{-st}sin(at))/(a^2+s^2)-(se^{-st}cos(at))/(a^2+s^2)+C